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If A=[1234]A = \begin{bmatrix}1 & 2\\3 & 4\end{bmatrix} such that A2+βA+γI=0A^2 + \beta A + \gamma I = 0 then β+γ\beta + \gamma is equal to:

(where II is an identity matrix of order 2)

Solution

✅ Correct Option: 3

By the Cayley-Hamilton theorem a 2×22\times2 matrix satisfies A2−(trace)A+(det⁡)I=0A^2 - (\text{trace})A + (\det)I = 0.

Here trace =5=5 and det⁡=−2\det = -2, so A2−5A−2I=0A^2 - 5A - 2I = 0, giving β=−5\beta = -5 and γ=−2\gamma = -2.

Hence β+γ=−7\beta + \gamma = -7.

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